In lab, Will burns a 0.6 g peanut beneath 48 g of water, which increases in temperature from 22 degree C to 55 degree C. The amount of heat absorbed by the water can be found with the equation Q =cm \Delta T, where Q is the amount of heat, c the specific heat of water, m the mass of water, and \Delta T the change in the water's temperature.

Question

In lab, Will burns a 0.6 g peanut beneath 48 g of water, which increases in temperature from 22 degree C to 55 degree C. The amount of heat absorbed by the water can be found with the equation Q =cm \Delta T, where Q is the amount of heat, c the specific heat of water, m the mass of water, and \Delta T the change in the water's temperature. 
A)Assuming that 38 % of the heat released makes its way to the water, find the food value of the peanut. 
B)What is the food value in calories per gram? 
C)What is the food value in Calories per gram?

bobpursley · Answer

heat from peanut= mc*deltaT/.38 value per gram= heat from peanut/.6